Definition: Metrics $d_1$ and $d_2$ on $X$ are topologically equivalent iff $d_1(x,x_n)\to 0 \iff d_2(x,x_n)\to 0$ for every $\{x_n\}\subset X$ and $x\in X$.
Definition: Metrics $d_1$ and $d_2$ on $X$ are metrically equivalent iff $d_1(x,y)\le ad_2(x,y)$ and $d_2(x,y)\le bd_1(x,y)$ for some $a,b\in\mathbb{R}$ and all $x,y\in X$.
I can easily see that metrically equivalent metrics are topologically equivalent.
I'm attempting to prove that $d_1$ and $d_2$ are topologically equivalent iff $(X,d_1)$ and $(X,d_2)$ have the same topology (collection of all open subsets).
Here's my (attempted) proof:
Let $U$ be open in $(X,d_1)$; i.e., $\forall x\in X$, $\exists r_1\in\mathbb{R}^+: d_1(x,y)<r_1 \implies y\in U$.
I'm trying to show that $U$ is open in $(X, d_2)$ by proving that $\exists r_2\in\mathbb{R}^+: d_2(x,y)<r_2\implies d_1(x,y)<r_1\implies y\in U$.
Assume the last statement doesn't hold: $\forall r_2>0, \exists y\in X: d_2(x,y)<r_2$ but $d_1(x,y)\ge r_1$.
Choose $\{x_n\}$ such that $d_2(x,x_n)<1/n$ but $d_1(x,x_n)\ge r_1$. So $d_2(x,x_n)\to 0$ but $d_1(x,x_n)\not\to 0$. Contradiction!
I'm concerned that there might exist $N$ such that $d_2(x,x_N)<1/N\implies x_N = x$. In that case, $d_1(x,x_N) = 0 < r_1$, still contradiction. Is this contradiction valid for my proof?
To prove the converse, that is, if $U$ open in $(X,d_2)$ implies $U$ open in $(X,d_1)$, then $d_1(x,x_n)\to 0$ implies $d_2(x,x_n)\to 0$; I could prove the identity mapping from $(X,d_1)$ to $(X,d_2)$ is continuous, then the preservation of convergence. Is there a direct approach to proving preservation of convergence?
My next question: Are topologically equivalent metrics metrically equivalent? I guess no. Can you give a counter example please?
Thank you!
Best Answer
Your argument seems kind of off: what is $y$ in your definition of the open set? It should probably start with $\forall x\in U\,\,\exists r_1\,\,\forall y\in X$. I did not carefully analyse the logic of your argument, but there are serious issues with the presentation: you seem to be a little sloppy with the symbolic treatment. If you absolutely need to write those things symbolically, you should take care to do it properly. Moreover, you seem to have fixed $x$ somewhere between the first and the second paragraphs of the proof. You should avoid doing this sort of thing without saying, at least until you become a more seasoned mathematician (and even then you should do it sparingly, if at all, especially in writing!).
The most direct way to prove the fact is probably by noticing that in a metric space, a set is closed if and only if it contains all the limits of (convergent) sequences of its elements. Once you have that, it is easy to see that your definition of topological equivalence makes the closed sets the same.
As for the counterexample, compare the standard Euclidean metric $d_E$ on the real line and $d'(x,y)=\arctan(d_E(x,y))$. They are topologically equivalent (easily), but not metrically equivalent. Even more trivially, you can take $d''(x,y)=\min(d_E(x,y),1)$. A bounded metric will never be metrically equivalent to an unbounded one.